As we know initially PF in case of induction motor is lower, but after some time PF becomes improve. Can any one explains that how we can improve the PF instead by using the parallel capacitor bank?
In starting due to higher magnetizing current component, the power factor is low. Could improve by starting motor, with gradually and slowly increasing voltage(auto transformer or proper starter), and after reaching rated speed, run it only at possible full load,....
I give recommendations for standard PF correction towards the end, But I started with just a little background information.
Power factor is a measure of how effectively the power provided is being used; it is the ratio of real power to apparent power.
Interesting question, why worry about PF?
Bad PF increases wire losses
Bad Power Factor increases infrastructure strain and heat on components
Reduces the amount of usable energy- Cost more to supply more energy to do the same work.
Why does this happen:
Inductive loads, like motors or transformer leakages, cause the current to lag behind voltage and drop the power factor. In most industrial settings, a lagging power factor is common.
So,
First establish:
cosφ= [active power/apparent power]
Power Factor is the cosine of the phase difference between the source voltage and current. It refers to the fraction of total power (apparent power) which is utilised to do the useful work called active power.
Also,
Real power is given by P = VIcosφ
A little more detail of why worry about PF:
The electrical current is inversely proportional to cosφ for transferring a given amount of power at a certain voltage. Which is one reason listed above for PF reduction. (Poor PF increases current flow)
There can also be PF contribution to voltage drop and poor voltage regulation on distribution and transmission lines. (It reduces the maximum useful power which can be transmitted along the line. The power which can be transmitted is proportional to the power factor. PF increases the percentage of losses along the line by a factor of 1/PF 2 1/PF2 where PF is the power factor.)
Also the KVA of a specific machine can be calculated as KVA=[KW/cosφ]
Generation plants need to be able to create reactive power to counterbalance poor power factors and also to ensure they can drive their output through the grid towards the loads.
1.) A large switched capacitor bank. As the load and load power factor change, capacitors would be switched in and out to match the reactive power required. Most large electric motors would already have capacitors built in.
3.)STATCOMs - static VAR (volt amp reactive) compensators. These are power electronics devices designed to supply reactive power which can react much quicker than switched capacitor banks.